台綜大轉學考微積分 108 B 卷解答

第一小題沒有陷阱，連續函數的極限值等於函數值。第二小題是不定型，可用羅必達法則，也可用泰勒展開式(須注意有無在收斂半徑內)。

〔1〕
$\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x}} = 1　∎$

〔2〕
法一 使用羅必達法則
$\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{e^x – 1 – x}}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{e^x – 1}}\mathop = \limits^{\rm{L}} \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{e^x }} = 1　∎$
法二 使用泰勒展開式
$\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{e^x – 1 – x}}{\rm{ = }}\mathop {\lim }\limits_{x \to 0} \frac{{1 – (1 – \frac{{x^2 }}{{2!}} + \frac{{x^4 }}{{4!}} – \frac{{x^6 }}{{6!}} + \cdots )}}{{(1 + x + \frac{{x^2 }}{{2!}} + \frac{{x^3 }}{{3!}} + \cdots ) – 1 – x}} = \frac{{\frac{1}{2}}}{{{\rm{ }}\frac{1}{2}{\rm{ }}}} = 1　∎$

附註：$\mathop = \limits^{\rm{L}}$ 表示此等號使用羅必達法則 (L’Hôpital’s rule)

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此題考點：羅必達法則、泰勒定理

本題要求半徑增加的速率，將體積與時間的關係列式後微分即可。

$V(t){\rm{ = }}\frac{4}{3}\pi \left[ {r(t)} \right]^3$
$\Rightarrow \underbrace {V'(t)}_{5cm^3 /s} = 4\pi \underbrace {\left[ {r(t)} \right]^2 }_{(2cm)^2 } \cdot r'(t)$

Let $r(t) = 2$
$\Rightarrow 5 = 16\pi \cdot r'(t) \Rightarrow r'(t) = \frac{5}{{16\pi }}　∎$

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此題考點：微分求極值

本題要求 $f(x)$ 的切線，已經有切點了，因此只要找出 $f'(x)$ 即可。

$\because x^2+(f(x))^2={\rm{(2}}x^2 {\rm{ + 2(}}f(x))^2 – x{\rm{)}}^2$
$\therefore 2x + 2f(x) \cdot f'(x) = 2(2x^2 + 2(f(x))^2 – x) \cdot (4x + 4f(x) \cdot f'(x) – 1)$

At $x=0$. $(x,f(x)) = (0,\frac{1}{2})$
$\Rightarrow 2 \cdot \frac{1}{2} \cdot f'(0) = 2 \cdot (2 \cdot \frac{1}{4})(4 \cdot \frac{1}{2} \cdot f'(0) – 1)$
$\Rightarrow f'(0) = 2f'(0) – 1$
$\Rightarrow f'(0) = 1$

Tangent line: $x-y=-\frac{1}{2}　∎$

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此題考點：切線專論

求 $f(x)$ 的極值需要微分，而 $f(x)$ 裡有積分式且上下界包含變數，想到微積分基本定理。

$\begin{array}{cl} f'(x) & =\frac{d}{{dx}}\int_1^{\sqrt x } {\frac{{e^t }}{t}} dt \\ & =\frac{{e^{\sqrt x } }}{{\sqrt x }} \cdot \frac{1}{2} \cdot x^{ – \frac{1}{2}} \\ & =\frac{{e^{\sqrt x } }}{{2x}} > 0\ \text{on }[1,\infty) \end{array}$

$\because \ f'(x)>0$ on $[1,\infty)$
$\therefore f'(x)$ increases on $[1,\infty)$
$\Rightarrow \mathop {\min }\limits_{x \in \left[ {1,\infty } \right)} f(x) = f(1) = \int_1^1 {\frac{{e^t }}{t}} dt = 0　∎$

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此題考點：微積分基本定理 II、微分求極值

本題並非旋轉體積分，因此回到積分的觀念下手；積分時，可以多多利用對稱的性質讓計算變簡單。

V $=2\int_0^1 {2(1 – x^2 )} dx$ (利用對稱性質算 $x \ge 0$ 的部分再乘 2)
V $=8\left[ {x – \frac{{x^3 }}{3}} \right]_{x = 0}^{x = 1}$
V $=8(1 – \frac{1}{3}) = \frac{{16}}{3}　∎$

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此題考點：定積分直觀觀念

四大積分法：變數變換、三角置換、分部積分和部分分式 是一定會考的內容，一定要熟練。

Let $u = \tan ^{ – 1} x$
$\begin{array}{cl} \int_0^\infty {\frac{{x\tan ^{ – 1} x}}{{(1 + x^2 )^2 }}} dx & =\int_0^{\frac{\pi }{2}} {\frac{{(\tan u) \cdot u}}{{(\underbrace {1 + \tan ^2 u}_{\sec ^2 u})}}} du \\ & =\int_0^{\frac{\pi }{2}} {\frac{{\sin u}}{{\cos u}} \cdot \cos ^2 u \cdot u} du \\ & =\frac{1}{2}\int_0^{\frac{\pi }{2}} {u \cdot \sin (2u)} du \\&=\frac{1}{2}\left[ {\left. {\frac{{ – u \cdot \cos (2u)}}{2}} \right|_{u = 0}^{u = \frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\frac{{\cos (2u)}}{2}} du} \right]\\&=\frac{1}{2}\left[ {\frac{{ – \frac{\pi }{2} \times ( – 1)}}{2} + \frac{1}{2} \cdot \left. {\frac{{\sin (2u)}}{2}} \right|_{u = 0}^{u = \frac{\pi }{2}} } \right]\\&=\frac{1}{2} \times \frac{\pi }{4} = \frac{\pi }{8}　∎ \end{array}$

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此題考點：三角置換法、分部積分

本題要求收斂半徑，這裡示範比較審斂法。

$r = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{( – 2)^{n + 1} }}{{(n + 1)^2 }}}}{{\frac{{( – 2)^n }}{{n^2 }}}}} \right|}} = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{2n^2 }}{{(n + 1)^2 }}} \right|}} = \frac{1}{2}$

$\Rightarrow $ the series converges on $( – \frac{1}{2},\frac{1}{2})$
For $x=\frac{1}{2}$
$\sum\limits_{n = 1}^\infty {\frac{{( – 2)^n }}{{n^2 }}x^n }=\sum\limits_{n{\rm{ = 1}}}^\infty {\frac{{( – 2)^n }}{{n^2 }} \cdot (\frac{1}{2})^n } = \sum\limits_{n = 1}^\infty {\frac{{( – 1)^n }}{{n^2 }}}$ converges. (by alternating series test)

For $x=-\frac{1}{2}$
$\sum\limits_{n = 1}^\infty {\frac{{( – 2)^n }}{{n^2 }} \cdot ( – \frac{1}{2})^n } = \sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}}$

$\Rightarrow $ interval of convergence: $\left[ { – \frac{1}{2},\frac{1}{2}} \right]　∎$

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此題考點：比較審斂法、根值審斂法

求(偏)微分遇到分段函數，有很高的機率需要使用定義處理。

$\begin{array}{cl} f_x (0,0) & =\mathop {\lim }\limits_{h \to 0} \frac{{f(h,0) – f(0,0)}}{h} \\ & =\mathop {\lim }\limits_{h \to 0} \frac{{\frac{0}{{h^2 }} – 0}}{h} = 0　∎\end{array}$

$\begin{array}{cl} f_y (0,0) & =\mathop {\lim }\limits_{k \to 0} \frac{{f(0,k) – f(0,0)}}{k} \\ & =\mathop {\lim }\limits_{k \to 0} \frac{{\frac{{\frac{1}{2}k^3 – 0}}{{k^2 }}}}{k} = \frac{1}{2}　∎ \end{array}$

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此題考點：二變數函數的偏微分

題目已有要求使用拉格朗日乘數法找極值，而這個方法很需要熟練的計算技巧，平時要多練習。

Let $g(x,y,z)=xy+2xz+2yz-12$
Let $\nabla f+\lambda \nabla g=0$
$\Rightarrow (yz,xz,xy) + \lambda (y + 2z,x + 2z,2x + 2y) = 0$
$\Rightarrow \left\{ {\begin{array}{cl}{yz + \lambda y + 2\lambda z = 0} & (1) \\{xz + \lambda x + 2\lambda z = 0} & (2) \\{xy + 2\lambda x + 2\lambda y = 0} & (3) \\ \end{array}} \right.$
◎◎ $\begin{array}{cl} xy+2xz+2yz=12 & (4) \end{array}$

$\begin{array}{cl} (1)-(2) & \Rightarrow z(y – x) + \lambda (y – x) = 0 \\ & \Rightarrow (y – x) \cdot (z + \lambda ) = 0 \Rightarrow x = y \text{ or }z=-\lambda \end{array}$
If $z=-\lambda$
$\mathop \Rightarrow \limits^{(1)} – \lambda y + \lambda y – 2\lambda ^2 = 0 \Rightarrow \lambda = 0 \Rightarrow z = 0$
$\mathop \Rightarrow \limits^{(2),(4)} \left\{ \begin{array}{l} xy = 0 \\{xy = 12}\end{array} ( \to \leftarrow ) \right. \Rightarrow z \ne \lambda$
$\Rightarrow x = y$
$\mathop \Rightarrow \limits^{(2),(3),(4)}\left\{ \begin{array}{ll} xz + \lambda x + 2\lambda z = 0 & (6) \\x^2 + 2\lambda x + 2\lambda x = 0 \Rightarrow x^2 + 4\lambda x = 0 & (5) \\ x^2 + 2xz + 2xz = 12 \Rightarrow x^2 + 4xz = 12 & (7)\end{array} \right.$

$\begin{array}{cl} (5) & \Rightarrow x(x + 4\lambda ) = 0 \\ & \Rightarrow x = 0 \text{ or }x=-4\lambda \end{array}$
$\because x^2+4xz=12$
$\therefore x \ne 0$
$\Rightarrow x=-4\lambda$
$\because x=y$
$\therefore x=4y=-4\lambda$
$\mathop \Rightarrow \limits^{(6),(7)}\left\{ \begin{array}{l} -4\lambda z – 4\lambda ^2 + 2\lambda z = 0\Rightarrow \lambda (2\lambda + z) = 0\\ \begin{array}{ll} 16\lambda^2-16\lambda z=12 & (8)\end{array} \end{array}\right.$
$\Rightarrow \lambda = 0\text{ or }z=-2\lambda$
If $\lambda = 0\mathop \Rightarrow \limits^{(8)} 0 = 12{\rm{ }}( \to \leftarrow ) \Rightarrow \lambda \ne 0$
$\Rightarrow z=-2\lambda$
$\mathop \Rightarrow \limits^{(8)} (4\lambda )^2 + 32\lambda ^2 = 12 \Rightarrow \lambda = \pm \frac{1}{2}$
$\Rightarrow (x,y,z) = ( – 4\lambda , – 4\lambda , – 2\lambda ) = ( – 2, – 2, – 1)$ or $(2,2,1)$

當 $(x,y,z) = ( – 2, – 2, – 1)$ 時，$f(x,y,z)=-4$
當 $(x,y,z)=(2,2,1)$ 時，$f(x,y,z)=4$
極值 $\left\{ {\begin{array}{l} {{\rm{Max}} = 4} \\ {{\rm{min}} = – 4} \\ \end{array}} \right.　∎$

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此題考點：拉格朗日乘數法

不易算的二重積分可嘗試極座標變換或交換積分次序(Fubini 定理)，本題被積函數有 $x^2+y^2$ 因此使用極座標變換。

Using polar coordinates:
Let $x = r\cos \theta$, $y = r\sin \theta$
$\Rightarrow x^2+y^2=r^2$, $dxdy=rdrd\theta$
$\int_{ – a}^a {\int_{ – \sqrt {a^2 – x^2 } }^{\sqrt {a^2 – x^2 } } {\cos (x^2 + y^2 )} } dydx=\int_0^{2\pi } {\int_0^a {\cos r^2 } } rdrd\theta$

Let $u=r^2\Rightarrow du=2rdr$
$\begin{array}{cl} \int_0^{2\pi } {\int_0^a {\cos r^2 } } rdrd\theta & =\int_0^{2\pi } {\int_0^{a^2 } {\cos u} \frac{{du}}{2}} d\theta \\ & =\frac{1}{2}\int_0^{2\pi } {\left. {\sin u} \right|_{u = 0}^{u = a^2 } } d\theta \\ & =\frac{1}{2}\int_0^{2\pi } {\sin a^2 } d\theta \\&=\pi \sin a^2　∎ \end{array}$

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此題考點：二重積分座標變換